How do you solve the equation #2(x+2)^2=72#?

4 Answers
Mar 22, 2017

Answer:

#"answer "{x=-8" , "x=4}#

Explanation:

#2(x+2)^2=72#

#"divide both sides by 2"#

#(cancel(2)(x+2)^2)/cancel(2)=(72)/2#

#(x+2)^2=36#

#"take square root both sides"#

#(x+2)=+-6#

#"if "x+2=-6" , then " x=-8#

#"if "x+2=6 " , then " x=4 #

Mar 22, 2017

Answer:

#x = 4# or #-8#

Explanation:

#2(x+2)^2 = 72#

divide by #2#:

#(x+2)^2 = 36#

square root:

#x+2 = 6# or #-6#

subtract #2#:

#x = 4# or #-8#

Mar 22, 2017

Answer:

#x={4,-8}#

Explanation:

First divide both sides by two to get #(x+2)^2=36#. Next take the square root of both sides to get #x+2=+-6#. Now split this into two equations for both the positive and negative six. These look like: #x+2=6# and #x+2=-6#. For both equations, subtract two from both sides which gives you #x=4# and #x=-8#. Hence #x={4,-8}#.

Mar 22, 2017

Answer:

#x=-8" or " x=4#

Explanation:

Divide both sides by 2

#rArr2/2(x+2)^2=72/2#

#rArr(x+2)^2=36#

Take the #color(blue)"square root of both sides"#

#sqrt((x+2)^2)=color(red)(+-)sqrt36#

#rArrx+2=color(red)(+-)6#

#• x+2=color(red)(6)#

subtract 2 from both sides.

#xcancel(+2)cancel(-2)=6-2#

#rArrx=4#

#• x+2=color(red)(-6)#

#rArrx=-8#

#color(blue)"as a check"#

Substitute these values into the left side of the equation and if equal to the right side then they are the solutions.

#x=4to2(4+2)^2=2xx36=72to" true"#

#x=-8to2(-8+2)^2=2xx36=72to" true"#

#rArrx=-8" or " x=4" are the solutions"#